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Friction Coefficient

  1. Oct 5, 2008 #1
    1. The problem statement, all variables and given/known data
    Two blocks are arranged at the ends of a mass-
    less string as shown in the figure. The system
    starts from rest. When the 1.81 kg mass has
    fallen through 0.336 m, its downward speed is
    1.25 m/s.
    The acceleration of gravity is 9.8 m/s2 .
    at is the frictional force between the
    3.13 kg mass and the table? Answer in units
    of N.

    3. The attempt at a solution
    m(table) T-friction=m(table)a
    m(fallen) T-m(fallen)g=m(fallen)a

    T=m(fallen)a-m(fallen)g
    T=9.3215

    T-friction=m(table)a
    -(coeff)N=m(table)a-T
    (coeff)N=T-m(table)a
    coeff=t-m(table)a
    -----------
    m(table)g
    =.1706

    Did I go about this correctly?
     
  2. jcsd
  3. Oct 6, 2008 #2

    Kurdt

    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Your answer is slightly off. I'd check your equation for the falling mass and work out you acceleration and tension again.
     
  4. Oct 6, 2008 #3
    i redid it and got .1376.
     
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